The Foundations of Chemistry

EXAMPLE 16-4 Method of Initial Rates

Use the following initial rate data to determine the form of the rate-law expression for the
reaction

3A
2B88n2C
D

Initial Initial Initial Rate of
Experiment [A] [B] Formation of D

1 1.00 10 ^2 M 1.00 10 ^2 M 6.00 10 ^3 Mmin^1

2 2.00 10 ^2 M 3.00 10 ^2 M 1.44 10 ^1 Mmin^1

3 1.00 10 ^2 M 2.00 10 ^2 M 1.20 10 ^2 Mmin^1

Plan

The rate law is of the form ratek[A]x[B]y. No two experiments have the same initial [B], so
let’s use the alternative method presented earlier to evaluate xand y.

Solution
The initial concentration of A is the same in experiments 1 and 3. We divide the third rate-
law expression by the corresponding terms in the first one



The initial concentrations of A are equal, so they cancel, as does k.Simplifying and then substi-
tuting known values of rates and [B],

 or 


y

2.0(2.0)y or y 1 The reaction is first order in [B].

No two of the experimental runs have the same concentrations of B, so we must proceed
somewhat differently. Let us compare experiments 1 and 2. The observed change in rate must
be due to the combinationof the changes in [A] and [B]. We can divide the second rate-law
expression by the corresponding terms in the first one, cancel the equal kvalues, and collect
terms.




x




y

Now let’s insert the known values for rates and concentrations and the known [B] exponent
of 1.




x




1

24.0(2.00)x(3.00)

8.00(2.00)x or x 3 The reaction is third order in [A].

The rate-law expression has the form ratek[A]^3 [B].

You should now work Exercise 22.

3.00 10 ^2 M



1.00 10 ^2 M

2.00 10 ^2 M



1.00 10 ^2 M

1.44 10 ^1 Mmin^1



6.00 10 ^3 Mmin^1

[B] 2



[B] 1

[A] 2



[A] 1

k[A] 2 x[B] 2 y



k[A] 1 x[B] 1 y

rate 2



rate 1

2.00 10 ^2 M



1.00 10 ^2 M

1.20 10 ^2 Mmin



6.00 10 ^3 Mmin

[B] 3 y



[B] 1 y

rate 3



rate 1

k[A] 3 x[B] 3 y



k[A] 1 x[B] 1 y

rate 3



rate 1

16-3 Concentrations of Reactants: The Rate-Law Expression 663